AM broadcast signal covering 995-1005 kHz radio frequency (RF) range is to be down-converted to an intermediate frequency (IF) range of 395-405 kHz where main amplification takes place in AM receivers. I need to find the mixing frequency.

i solved it this way

fm-1005=405 .     fm=1410
fm-995=405        fm=1400
fm-1005=395 .      fm=1400
fm-995=395 .      fm=1390

so the mixing frequency is 1400. I was just wondering if that the way of how to find the mixing frequency?

  • 1
    $\begingroup$ Please explain exactly what you're asking for. You've stated something that sounds like an exam question and your solution, but not what kind of answers you want from us. $\endgroup$
    – Kevin Reid AG6YO
    Jan 17, 2019 at 22:05
  • $\begingroup$ I was just wondering if that the way of how to find the mixing frequency? @KevinReidAG6YO $\endgroup$
    – natasha
    Jan 17, 2019 at 22:30
  • $\begingroup$ Please edit your question to say that, then. It's also good to ask a question that has an answer more complex than "Yes, you got it right." $\endgroup$
    – Kevin Reid AG6YO
    Jan 17, 2019 at 22:32
  • $\begingroup$ did I got it right? i am discussing my idea to solve it @KevinReidAG6YO $\endgroup$
    – natasha
    Jan 17, 2019 at 22:39
  • 2
    $\begingroup$ That's sufficient. But when asking future questions, please try to make them more substantial than "am I doing this right?" For example, ask a specific question about the part of the theory you aren't confident about. In general, people will like your questions better and you'll get more informative answers if you post a more substantial question, that's about the fundamental concepts or a practical problem rather than how to answer an exam/homework question. $\endgroup$
    – Kevin Reid AG6YO
    Jan 17, 2019 at 22:51

1 Answer 1


Your answer in your question is not correct.

This question is solved in the same way as the last question you asked. The formulas are the same:

$$f_s=f_1+f_2 \tag 1$$ $$f_d=f_1-f_2 \tag 2$$

This time, the sum and difference is called IF (intermediate frequency) - a common term in mixer circuits. Then rearrange to find the LO (local oscillator) frequency:

$$LO=| IF-Signal | \tag 3$$ $$LO=IF+Signal \tag 4$$

Start with the low end of both ranges and plug them into the equations. There are two possible answers for the LO frequency.

You can then plug in the upper frequencies of the ranges in the same way to see that it works out to the same LO frequency.

Spoiler alert! The answer follows:

$$LO=| IF-Signal | = | 395 - 995 | = 600 \text{ kHz}$$

$$LO=IF+Signal = 395 + 995 = 1390 \text{ kHz}$$

The issue with using the 1390 kHz LO is that the IF passband will be inverted compared to the original signal. Consider that 1390 kHz - 995 kHz = 395 kHz but 1390 - 1005 kHz = 385 kHz, so the frequency image is inverted. As a result, the correct LO is 600 kHz.

  • $\begingroup$ 395=fm+995 fm=600 405=fm+1005 fm=600 for eq. 3 and 4 what is the signal values? $\endgroup$
    – natasha
    Jan 18, 2019 at 0:10
  • 1
    $\begingroup$ @natasha The AM signal: 995 to 1005 kHz. $\endgroup$
    – Glenn W9IQ
    Jan 18, 2019 at 0:15
  • $\begingroup$ So LO = 395,1605 $\endgroup$
    – natasha
    Jan 18, 2019 at 0:21
  • 1
    $\begingroup$ @natasha Use just the low end of the frequencies to compute the answers. Or use just the high end of both for the same result. $\endgroup$
    – Glenn W9IQ
    Jan 18, 2019 at 0:27
  • $\begingroup$ 395,1595 if I calculate LO=IF+Sig, LO=IF-SIG it it will not be the same if I put the high frequencies of the range $\endgroup$
    – natasha
    Jan 18, 2019 at 0:55

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